QJEEM24Phb7a1: JEE Main 2024 Physics - Class 11 JEEM24Phb7a1 Solution: A diatomic gas does of work in an isobaric expansion. The heat…

To find the heat given to a diatomic gas during an isobaric (constant pressure) expansion, we can use the formula that relates the work done by the gas, the heat added to the system, and the change in the internal energy of the system. The first law of thermodynamics states that:

$\Delta Q = \Delta U + W$

where:

• $\Delta Q$ is the heat added to the system,
• $\Delta U$ is the change in internal energy of the system, and
• $W$ is the work done by the gas.

For an isobaric process, the work done $W$ is given by:

$W = P \Delta V$

We're given that $W = 100 \ \mathrm{J}$ for this process, so:

$W = 100 \ \mathrm{J}$

The change in internal energy $\Delta U$ for an ideal gas can also be related to the temperature change and the specific heat capacity at constant volume $C_v$ . Using the equation:

$\Delta U = nC_v\Delta T$

However, without direct values for $n$ , $\Delta T$ , or $C_v$ , we need to rely on the relation between the provided work and the heat capacity ratio $\gamma$ to find the heat added. For a diatomic gas, $\gamma = C_p/C_v$ . The heat added at constant pressure can also be described as:

$\Delta Q = nC_p\Delta T$

Since we know that for an ideal gas, the work done on the gas during an isobaric process is related to the heat added by the ratio of the specific heats ( $\gamma$ ), we can use the fact that $C_p = \gamma C_v$ , and relating that to the work done, we get:

$\Delta Q = \frac{\gamma}{\gamma - 1} W$

Substituting the known values, with $\gamma = 1.4$ , and $W = 100 \ \mathrm{J}$ , we find:

$\Delta Q = \frac{1.4}{1.4 - 1} \times 100 \ \mathrm{J} = \frac{1.4}{0.4} \times 100 \ \mathrm{J} = 3.5 \times 100 \ \mathrm{J} = 350 \ \mathrm{J}$

Therefore, the heat given to the gas is $\Delta Q = 350 \ \mathrm{J}$ , so the correct option is:

Option C

350 J